A vernier calipers have $1 m m$ marks on the main scale. It has $20$ equal divisions on the vernier scale which matches with $16$ main scale divisions. For this vernier calipers, the least count is :

0.02 m m

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0.05 m m

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0.1 m m

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0.2 m m

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Solution

The correct option is D $0.2 m m$
Least count, $L . C = 1 M . S . D - 1 V . S . D$
As vernier scale division equals matches on main scale divison, therefore
$L . C = (1 - \frac{16}{20}) M . S . D$
$= (1 - 0.8) (1 m m) = 0.2 m m$

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A vernier calipers have 1mm marks on the main scale. It has 20 equal divisions on the vernier scale which matches with 16 main scale divisions. For this vernier calipers, the least count is :

The correct option is D 0.2mmLeast count, L.C=1M.S.D−1V.S. DAs vernier scale division equals matches on main scale divison, therefore L.C=(1−1620)M.S.D =(1−0.8)(1mm)=0.2mm

A vernier calipers have $1 m m$ marks on the main scale. It has $20$ equal divisions on the vernier scale which matches with $16$ main scale divisions. For this vernier calipers, the least count is :

The correct option is D $0.2 m m$
Least count, $L . C = 1 M . S . D - 1 V . S . D$
As vernier scale division equals matches on main scale divison, therefore
$L . C = (1 - \frac{16}{20}) M . S . D$
$= (1 - 0.8) (1 m m) = 0.2 m m$